Find Nth Derivative of Square Root of X

[dnt]

how do you find the general formula for the nth derivative of x^1/2 (square root of x)?

i can get the full formulas but i cannot put it together as a general formula:

n=1 (1st der): (1/2)(x^-1/2)
n=2 (2nd der): (1/2)(-1/2)(x^-3/2)
n=3 (3rd der): (1/2)(-1/2)(-3/2)(x^-5/2)
n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2)
...
nth der = ?

i have it right so far, correct? i just can't figure out how put it in terms of n (is it an arithmetic series or something)?

thanks.

 

[Data]

Try to figure out a formula that works for the terms you have (think factorials). Then prove by induction that it works for all n.

 

[dnt]

i think the denominator will be 2^n

i think the exponent of x will be [-(2n-1)]

but i can't figure out the numerator part that goes 1,-1,-3,-5

would that be 3-2n?

 

[dnt]

(3-2n)/2^n * x^(1-2n)

does that work?

 

[0rthodontist]

n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2)

= (-1)^3 \frac{(1)(3)(5)}{2^4} \cdot x^{-\frac{7}{2}}
= (-1)^3 \frac{(1)(2)(3)(4)(5)(6)}{(1)(2)(3)(2^7)} \cdot x^{-\frac{7}{2}}
= (-1)^3 \frac{6!}{(3!)(2^7)} \cdot x^{-\frac{7}{2}}

By the way, instead of asking us if it works, why not find out yourself? Also the exponent of n is not what you said. It starts at 1/2 and decreases by 1 each time.

 

[dnt]

but how does that work in terms of n?

 

[0rthodontist]

Do some guessing and checking to figure out how it is generalized.

 

[dnt]

so is my previous guess of (3-2n)/2^n * x^(1-2n) totally wrong?

 

[0rthodontist]

It is, but you should verify that yourself.

 

[Data]

Well, do any of the derivatives you've found so far have integer powers of x (as your guess would predict)? Does your guess work to find the derivatives you've already calculated?

I'll give you a hint: yes you'll need powers of 2 in the denominator. Also look at the numerators of the coefficients. You're multiplying together a lot of numbers to get it in each case. Do you see a pattern in the numbers you're mulplying together?

 

[HallsofIvy]

Also, your statement that "the numerator part goes 1,-1,-3,-5" is wrong.

What you put in your very first post was:
n=1 (1st der): (1/2)(x^-1/2) = (1/2)(x^(-1/2))
n=2 (2nd der): (1/2)(-1/2)(x^-3/2) = (-1/4)(x^(-3/2))
n=3 (3rd der): (1/2)(-1/2)(-3/2)(x^-5/2) = (+3/8)(x^(-5/2))
n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2) = (-15/16)(x^(-7/2))
(I've added the last in each line.)
so the numerators are 1, -1, 3= 1*3, -15= -1*3*15.

It might help you to realize this:

A product of even integers is 2*4*6*...*2n= (2*1)(2*2)(2*3)...(2*n)= (2*2*2...2)(1*2*3*...n)= 2ⁿn!.

A product of odd integer, 1*3*5*7*...*(2n+1) is missing the even integers: multiply and divide by them:
\frac{1*2*3*4*5*...*(2n)*(2n+1)}{2*4*6*...*(2n)}= \frac{(2n+1)!}{2^n(n!)}

 

[dnt]

i didnt know that HoI. thanks. it also didnt help that our teacher told us not to multiply the stuff out to look for patterns...

however, that still doesn't explain how the numerator starts 1,-1...if i use what you did i start at 3/8 (which is the 3rd derivative, not the first)

 

[dnt]

is the 2nd part x^(n-1/2)? am i at least correct there?

 

[0rthodontist]

Your teacher gave you bad advice if he/she meant not to try any examples--it is good to multiply it out, check, look for a pattern, generalize. For one thing it's easier to deal with say the 4th derivative than with an arbitrary product notation or ellipsis. Though your final solution and derivation shoud be in the general form, it's often easier to work it out first for yourself on a specific example.

You should be able to tell yourself whether or not x^n-1/2 is correct.

 

[lapo3399]

Wouldn't it be ((3-2n)/2^n)*x^(1/2 - n)?

 

[dnt]

ok i tried using the equation for the product of odd in integers, which started me at 3 if i used n=1.

so i backed it up by substituting (n-1) for n to get:

(2n-1)!/[2^(n-1)*(n-1)!]

but that only brings me to the 2nd derivative if i plug in n=1.

how can i back it up another spot so that n=1 gives me the 1st derivative? i tried substituting n-2 for n but that gives me a negative factorial on the numerator.

i just cannot figure out the pattern for 1,1,3,15,105,etc...i know its the product of odd integers but those first numbers screw me up. how do i get it to start 1,1... instead of 1,3...

 

[dnt]

Your teacher gave you bad advice--it is good to multiply it out, check, look for a pattern, generalize. For one thing it's easier to deal with say the 4th derivative than with an arbitrary product notation or ellipsis.

You should be able to tell yourself whether or not x^n-1/2 is correct.

oops i meant to make it negative:

x^1/2-n

i think that part is correct but the rest is tough

 

[dnt]

Wouldn't it be ((3-2n)/2^n)*x^(1/2 - n)?

dont think that works because eg, n=4 gives you:

(-5/16) for the first part when it should be -15/16

 

[0rthodontist]

ok i tried using the equation for the product of odd in integers, which started me at 3 if i used n=1.

so i backed it up by substituting (n-1) for n to get:

(2n-1)!/[2^(n-1)*(n-1)!]

but that only brings me to the 2nd derivative if i plug in n=1.

how can i back it up another spot so that n=1 gives me the 1st derivative? i tried substituting n-2 for n but that gives me a negative factorial on the numerator.

i just cannot figure out the pattern for 1,1,3,15,105,etc...i know its the product of odd integers but those first numbers screw me up. how do i get it to start 1,1... instead of 1,3...

It looks okay the way it is, now you just need to include the powers of -1 (sign alternation), the additional denominator of 2^(n), and the power of x, which you already have correctly.

 

[dnt]

i still don't see why that's ok.

using this: (2n-1)!/[2^(n-1)*(n-1)!]

gives me 1,3,15,105 for n=1,2,3,4

which means it starts off with the 2nd derivative for n=1.

it should start with the first derivative for n=1

 

[0rthodontist]

Oh, I see what you mean. So just use n-2 and treat the first derivative as a separate case. That formula for product of odd integers is anyway only good when the integers are all positive.

 

[Data]

Halls' expression isn't quite what you need. Keep in mind that all three of these expressions are for the product of the first n+1 odd numbers, and for your derivatives you want the product of the first n-1 odd numbers.

(1)(3)(5)...(2n+1) = \frac{(2n+1)!}{n!2^n} = \frac{(2n+2)!}{(n+1)!2^{n+1}}Give the expression on the farthest right a try :)

(the reason that Halls' doesn't work is that it's only applicable for n \geq 0, whereas the one on the right works for n \geq -1 - in the case n=-1, you get 1 [instead of the "product of the first 0 odd numbers" which doesn't make much sense])

 

[dnt]

still doesn't work.

n=1 gives 4!/(2!*4)=3

or am i suppose to substitute n-1 in for n?

 

[Data]

Take a look at what I said in the second sentence. That expression is the product of the first n+1 odd numbers and you want the product of the first n-1 odd numbers. So what should you sub in for n?

 

[dnt]

n-1 substituted in doenst work either! i still get 1,3,15,105 but i cannot get the first two terms to be 1,1.

this is impossible. (just kidding but its freaking hard!)

 

[Data]

try n-2

 

[dnt]

Take a look at what I said in the second sentence. That expression is the product of the first n+1 odd numbers and you want the product of the first n-1 odd numbers. So what should you sub in for n?

do i substitute n-2?

 

[Data]

try it! and if it works, then try to understand how

 

[dnt]

try n-2

ill try it. the post before you i typed in before your newest post :)

 

[dnt]

holy crap i think that works

 

[Data]

indeed

so can you explain why it works? (take a look at the bottom of the post where I suggested it)

 

[dnt]

ok so for my final answer:

[(-1)^(n+1)(2n-2)!]/[(2^n)(n-1)!(2^(n-1))]*X^(.5-n)

 

[0rthodontist]

Data's expression is the same as HallsofIvy's.

 

[dnt]

indeed

so can you explain why it works? (take a look at the bottom of the post where I suggested it)

still trying to understand it. i don't know what halls is though.

 

[Data]

yes, it's the same except that you can still evaluate it for n= -1 (even though the "product of the first 0 odd integers" doesn't make any sense). It happens to give dnt exactly what he needs though.

 

[dnt]

Data's expression is the same as HallsofIvy's.

your right...why didnt it work the first time i tried it?

dont know what i did. i think i substituted wrong or something.

 

[Data]

Halls is HallsofIvy

 

[dnt]

lol. i was thinking halls was some calc theorem :)

 

[Data]

occasionally an apparently useless thing (like, say, multiplying by 1, which is all I did) can allow you to avoid splitting things up into ugly "special cases"

 

[0rthodontist]

yes, it's the same except that you can still evaluate it for n= -1 (even though the "product of the first 0 odd integers" doesn't make any sense). It happens to give dnt exactly what he needs though~

Ah, good idea.
I do think it would have been easier to generalize from example (continuing from before, at the fourth derivative):
(-1)^3 \frac{6!}{(3!)(2^7)} \cdot x^{-\frac{7}{2}}
Now you know that you're going to change the sign every time, so the exponent on -1 is going to increase by 1 every time. The 6! came from 5 * 3 * 1 so that is going to increase by 2 every time. The 3! will increase by 1 every time, and the exponent on the 2 will increase by 1 every time from 5 * 3 * 1 and by another 1 every time from the denominator. So you can then write
(-1)^{n - 1} \frac{(2 * (n - 1))!}{((n-1)!)2^{2n - 1}} \cdot x^{\frac{1}{2} - n}
where all I did was make sure the various terms increase as they should, and otherwise adjust them by constants so they match the example. This saves you some fiddling with off-by-one errors and then you can verify it with more general notation.

 

[dnt]

Also, your statement that "the numerator part goes 1,-1,-3,-5" is wrong.

What you put in your very first post was:
n=1 (1st der): (1/2)(x^-1/2) = (1/2)(x^(-1/2))
n=2 (2nd der): (1/2)(-1/2)(x^-3/2) = (-1/4)(x^(-3/2))
n=3 (3rd der): (1/2)(-1/2)(-3/2)(x^-5/2) = (+3/8)(x^(-5/2))
n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2) = (-15/16)(x^(-7/2))
(I've added the last in each line.)
so the numerators are 1, -1, 3= 1*3, -15= -1*3*15.

It might help you to realize this:

A product of even integers is 2*4*6*...*2n= (2*1)(2*2)(2*3)...(2*n)= (2*2*2...2)(1*2*3*...n)= 2ⁿn!.

A product of odd integer, 1*3*5*7*...*(2n+1) is missing the even integers: multiply and divide by them:
\frac{1*2*3*4*5*...*(2n)*(2n+1)}{2*4*6*...*(2n)}= \frac{(2n+1)!}{2^n(n!)}

i konw the problem has been solved but i wanted to make sure i really understood it and one more question about the above equation popped into my head.

in regards to that numerator (product of the first n integers) how does it equal (2n+1)! ?

eg. if n=5 (product of first 5 integers = 1*2*3*4*5), well that should be equal to 120 which is n!

if you use (2n+1)! you get 11!, a much bigger number. can someone reexplain that? thanks.

 

[dnt]

i konw the problem has been solved but i wanted to make sure i really understood it and one more question about the above equation popped into my head.

in regards to that numerator (product of the first n integers) how does it equal (2n+1)! ?

eg. if n=5 (product of first 5 integers = 1*2*3*4*5), well that should be equal to 120 which is n!

if you use (2n+1)! you get 11!, a much bigger number. can someone reexplain that? thanks.

can anyone explain it?
thanks.

 

[Data]

in regards to that numerator (product of the first n integers) how does it equal (2n+1)! ?

it doesn't. (2n+1)! is the product of the first 2n+1 integers, of course, not the first n integers.

But (2n+1)!/(n!2^n) is the product of the first n+1 odd integers, because you're multiplying together the first 2n+1 integers and then dividing out the even ones. Here n!2^n is the product of the first n even integers:

n!2^n = n(n-1)...(1)*2^n = (2n)(2n-2)...(2).

So in (2n+1)!/(n!2^n), you're multiplying together the first 2n+1 integers in the top, then you divide out the first n even integers in the bottom, leaving you with the first (2n+1)-n = n+1 odd integers!